A) 0
B) \[\frac{1}{2}{{(\alpha -\beta )}^{2}}\]
C) \[\frac{{{a}^{2}}}{2}{{(\alpha -\beta )}^{2}}\]
D) None of these
Correct Answer: C
Solution :
[c] \[\underset{x\to \alpha }{\mathop{\lim }}\,\frac{1-\cos (a{{x}^{2}}+bx+c)}{{{(x-\alpha )}^{2}}}\] \[=\underset{x\to \alpha }{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\left( \frac{a{{x}^{2}}+bx+c}{2} \right)}{{{(x-\alpha )}^{2}}}\] \[=\underset{x\to \alpha }{\mathop{\lim }}\,2{{\left[ \frac{\sin \,\frac{a(x-\alpha )(x-\beta )}{2}}{\frac{a(x-\alpha )(x-\beta )}{2}} \right]}^{2}}\times \frac{{{a}^{2}}{{(x-\beta )}^{2}}}{4}\] \[=\frac{{{a}^{2}}}{2}{{(\alpha -\beta )}^{2}}\] [using \[a{{x}^{2}}+bx+c=a(x-\alpha )(x-\beta )\]]
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